从原点开始不访问(X，Y)即可到达矩阵(M，N)的方式数

给定四个正整数M，N，X和Y ，任务是计算从大小矩阵的左上角(即(0，0) )到右下角(M，N)到达的所有可能方式(M + 1)x(N + 1)，而无需访问单元格(X，Y) 。假设从每个像元(i，j)您只能向右(i，j + 1)或向下(i + 1，j)移动。
例子：

Input: M = 2, N = 2, X = 1, Y = 1 
Output: 2 
Explanation: 
There are only 2 ways to reach (2, 2) without visiting (1, 1) and the two paths are: 
(0, 0) -> (0, 1) -> (0, 2) -> (1, 2) -> (2, 2) 
(0, 0) -> (1, 0) -> (2, 0) -> (2, 1) -> (2, 2)
Input: M = 5, N = 4, X = 3, Y = 2 
Output: 66 
Explanation: 
There are 66 ways to reach (5, 4) without visiting (3, 2). 
 

方法：
为了解决上述问题，我们的想法是从(0，0)减去(X，Y)到达的次数，然后通过访问(X， Y)从(X，Y)到达(M，N) 。 Y)从(0，0 )到达(M，N)的总数。
所以，

1. 从原点(0，0 )到(M，N)的到达方式数为：

$\text{Total ways from (0, 0) to (M, N)} = \binom{M + N}{M}$

1. 仅通过访问(X，Y)到达(M，N)的方式数是从(0，0 )到达(X，Y) ，然后给出从(X，Y)到达(M，N)的方式经过：

$\text{Total ways from (0, 0) to (X, Y)} = \binom{X + Y}{X}$
$\text{Total ways from (X, Y) to (M, N)} = \binom{M + N - X - Y}{M - X}$
 

Therefore, 
 

$\text{Total ways from (0, 0) to (M, N) only by visting (X, Y)} = (\binom{X + Y}{X}) * (\binom{M + N - X - Y}{M - X})$
 

1. 因此，方法总数的等式为：

$\binom{M + N}{M} - (\binom{X + Y}{X}) * (\binom{M + N - X - Y}{M - X})$

下面是上述方法的实现：

C++

// C++ program from the above approach
#include 
using namespace std;
 
int fact(int n);
 
// Function for computing nCr
int nCr(int n, int r)
{
    return fact(n)
           / (fact(r) * fact(n - r));
}
 
// Function to find factorial of a number
int fact(int n)
{
    int res = 1;
 
    for (int i = 2; i <= n; i++)
        res = res * i;
 
    return res;
}
 
// Function for counting the number
// of ways to reach (m, n) without
// visiting (x, y)
int countWays(int m, int n, int x, int y)
{
    return nCr(m + n, m)
           - nCr(x + y, x) * nCr(m + n
                                     - x - y,
                                 m - x);
}
 
// Driver Code
int main()
{
    // Given Dimensions of Matrix
    int m = 5;
    int n = 4;
 
    // Cell not to be visited
    int x = 3;
    int y = 2;
 
    // Function Call
    cout << countWays(m, n, x, y);
    return 0;
}

Java

// Java program from the above approach    
import java.util.*;    
class GFG{   
     
// Function for computing nCr    
public static int nCr(int n, int r)        
{    
    return fact(n) / (fact(r) * fact(n - r));        
}    
         
// Function to find factorial of a number    
public static int fact(int n)    
{    
    int res = 1;
     
    for(int i = 2; i <= n; i++)        
        res = res * i;        
    return res;        
}    
         
// Function for counting the number        
// of ways to reach (m, n) without        
// visiting (x, y)        
public static int countWays(int m, int n,
                            int x, int y)        
{    
    return nCr(m + n, m) -
           nCr(x + y, x) *
           nCr(m + n - x - y, m - x);        
}
 
// Driver code
public static void main(String[] args)
{    
     
    // Given Dimensions of Matrix    
    int m = 5;        
    int n = 4;        
             
    // Cell not to be visited    
    int x = 3;        
    int y = 2;        
             
    // Function Call    
    System.out.println(countWays(m, n, x, y));    
}    
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program for the above approach
 
# Function for computing nCr
def nCr(n, r):
     
    return (fact(n) // (fact(r) *
                        fact(n - r)))
 
# Function to find factorial of a number
def fact(n):
     
    res = 1
    for i in range(2, n + 1):
        res = res * i
 
    return res
 
# Function for counting the number
# of ways to reach (m, n) without
# visiting (x, y)
def countWays(m, n, x, y):
     
    return (nCr(m + n, m) - nCr(x + y, x) *
            nCr(m + n - x - y, m - x))
 
# Driver Code
 
# Given dimensions of Matrix
m = 5
n = 4
 
# Cell not to be visited
x = 3
y = 2
 
# Function call
print(countWays(m, n, x, y))
 
# This code is contributed by sanjoy_62

C#

// C# program from the above approach    
using System;
 
class GFG{
     
// Function for computing nCr    
public static int nCr(int n, int r)        
{    
    return fact(n) / (fact(r) * fact(n - r));        
}    
         
// Function to find factorial of a number    
public static int fact(int n)    
{    
    int res = 1;
     
    for(int i = 2; i <= n; i++)        
        res = res * i;
         
    return res;        
}    
         
// Function for counting the number        
// of ways to reach (m, n) without        
// visiting (x, y)        
public static int countWays(int m, int n,
                            int x, int y)        
{    
    return nCr(m + n, m) -
           nCr(x + y, x) *
           nCr(m + n - x - y, m - x);        
}
 
// Driver code
public static void Main(String[] args)
{    
     
    // Given dimensions of Matrix    
    int m = 5;        
    int n = 4;        
             
    // Cell not to be visited    
    int x = 3;        
    int y = 2;        
             
    // Function call    
    Console.WriteLine(countWays(m, n, x, y));    
}    
}
 
// This code is contributed by Rajput-Ji

Javascript

 
 
// Javascript Program to implement
// the above approach
 
// Function for computing nCr    
function nCr(n, r)        
{    
    return fact(n) / (fact(r) * fact(n - r));        
}    
           
// Function to find factorial of a number    
function fact(n)    
{    
    let res = 1;
       
    for(let i = 2; i <= n; i++)        
        res = res * i;        
    return res;        
}    
           
// Function for counting the number        
// of ways to reach (m, n) without        
// visiting (x, y)        
function countWays(m, n, x, y)        
{    
    return nCr(m + n, m) -
           nCr(x + y, x) *
           nCr(m + n - x - y, m - x);        
}
 
// Driver Code
     
    // Given Dimensions of Matrix    
    let m = 5;        
    let n = 4;        
               
    // Cell not to be visited    
    let x = 3;        
    let y = 2;        
               
    // Function Call    
    document.write(countWays(m, n, x, y));
 
// This code is contributed by avijitmondal1998.

输出：

66